mth 174 problem 7.3.3 (previously 7.3.15) x^4 e^(3x) **** what it is your antiderivative?
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20:35:16 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The integral of x^4e^(3x) we use #14 in the table where p(x)=x^4 p=4x^3 ax=3x The formula is 1/ap(x)e^(ax)-1/a * integral (p(x)e^(ax) dx 1/3x^4e^(3x) 1/3 *integral(4x^3 e^(3x) dx 1/3 x^4e^(3x) 4/3 *integral(x^3 e^3x) Use # 14 again p(x) = x^3 p(x)=2x ax=3x 1/3 x^4e^(3x) 4/3[1/3x^3e^(3x) -1/3* integral(3x^2e^(3x)] 1/3x^4e^(3x)-4/9x^3e^(3x)-4/3 *integral(x^2e^(3x) Use #14 again p(x) = x^2 p=2x ax=3x 1/3x^4e^(3x)-4/9x^3e^(3x)-4/3[1/3x^2e^(3x)-1/3 *integral(2xe^(3x) dx) Use #14 again p(x)=x p=1 ax=3x 1/3x^4e^(3x)-4/9x^3e^(3x)-4/9x^2e^(3x)-8/9*integral(xe^(3x) )dx 1/3x^4e^(3x)-4/9x^3e^(3x)-4/9x^2e^(3x)-8/9[1/3xe^(3x) -1/3 *integral e^(3x) dx 1/3x^4e^(3x) 4/9x^3e^(3x) 4/9x^2e^(3x)-8/27xe^(3x) 8/81 e^(3x) +c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The integral is of x^4 e^(3 x). x^4 is a polynomial, and e^(3 x) is of the form e^(a x). So the integrand is of the form p(x) e^(a x) with p(x) = x^4 and a = 3. The correct formula to use is #14 We obtain p ' (x) = 4 x^3 p '' (x) = 12 x^2 p ''' (x) = 24 x p '''' (x) = 24. Thus the solution is 1 / a * p(x) e^(a x) - 1 / a^2 * p ' (x) e^(a x) + 1 / a^3 * p''(x) e^(a x) - 1 / a^4 * p ''' (x) e^(a x) + 1 / a^5 * p''''(x) e^(a x) = 1 / 3 * x^4 e^(3 x) - 1 / 3^2 * 4 x^3 e^(3 x) + 1 / 3^3 * 12 x^2 e^(3 x) - 1 / 3^4 * 24 x e^(3 x) + 1 / 3^5 * 24 e^(3 x) = ((1/3) x^4 - (4/9) x^3 + (4/9) x^2 - (8/27)x + (8/81) e^(3x) + C
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20:35:18
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###I didnt factor out e^(3x) but the answer is the same otherwise. ------------------------------------------------ Self-critique Rating: ********************************************* Question: Which formula from the table did you use?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used #14 4 times confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You should have used formula 14, with a = 3 and p(x) = x^4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: Question 7.3 Problem 7 problem 7.3.7 (previously 7.3.33 1 / [ 1 + (z+2)^2 ) ]) **** What is your integral? **** Which formula from the table did you use and how did you get the integrand into the form of this formula?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Integral of 1/(1+(z+2)^2) We begin by letting u = z+2 So that du = dz So substitute in Integral 1/(1+u^2) At this point we know that this is the derivative of the arctangent so our answer would be Arctan u +c which is arctan(z+2)+c If we need to use the table we would use #24 which says 1/(x^2+a^2) =1/a arctan x/a + c We would left x=u and a=1 because a is a constant so substituting in we get 1/1arctan u/1 +c Arctan u +c Arctan(z+2) +c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you let u = x+2 then du = dz and the integrand becomes 1 / (1 + u^2). This is the derivative of arctan(u), so letting u = z+2 gives us the correct result arctan(z+2) + C Applying the formula: z is the variable of integration in the given problem, x is the variable of integration in the table. a is a constant, so a won't be z + 2. By Formula 24 the antiderivative of 1 / (a^2 + x^2) is 1/a * arcTan(x/a). Unlike some formulas in the table, this formula is easy to figure out using techniques of integration you should already be familiar with: 1 / (a^2 + x^2) = 1 / (a^2( 1 + x^2/a^2) ) = 1/a^2 ( 1 / (1 + (x/a)^2). Let u = x / a so du = dx / a; you get integrand 1 / a * (1 / (1 + u^2) ) , which has antiderivative 1/a * arcTan(u) = 1/a * arcTan(x/a). You don't really need to know all that, but it should clarify what is constant and what is variable. Starting with int(1/ (1+(z+2)^2) dz ), let x = z + 2 so dx = dz. You get int(1/ (1+x^2) dx ), which is formula 24 with a = 1. The result is 1/1 * arcTan(x/1), or just arcTan(x). Since x = z + 2, the final form of the integral is arcTan(z+2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: Problem 7.4 Problem 1 7.4.1 (previously 7.4.6). Integrate 2y / ( y^3 - y^2 + y - 1)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Integral 2y/(y^3-y^2+y-1) We see that the denominator has 4 terms so factor by grouping working with denominator we have y^2(y-1)+1(y-1) so rewriting the denominator and bringing the 2 out front we have 2*integral y/[(y^2+1)(y-1) So now we use partial fractions and will remember to deal with the 2 out front at the end Ay+b/(y^2+1) + c/(y-1) =y/[(y^2+1)(y-1)] So to get a common denominator we multiply the first term by (y-1) and the second term by (y^2+1) So we have for numerator only (ay+b)(y-1) + c(y^2+1)=y Factoring we get ay^2-ay+by-b+cy^2+c=1y+0 Y^2(a+c) + y(-a+b) +1(c-+b) = 1y+0 Since the middle term has the y we set a+b=1 and get b=1+a for the other terms we set equal to 0 so we have a+c=0 a=-c c-b=0 c=b now we have a system of equations to solve 1+a=b 1+a=c 1+a=-a 1=-2a A=-1/2 A=-c so c=1/2 C=b so b=1/2 Now we go back to 2*integral ay+b/(y^2+1) + c/(y-1) and substitute in values 2*integral -1/2y + 1/2/(y^2+1) + 2*integral .5/(y-1) Working on first integral factor out -.5 2*-.5 = -1 so for the first integral we have -1*integral (y-1)/(y^2+1) Let u = y-1 du = 1dy So for the first integral we have -1*integral u/(y^2+1) which we recognize as u(arctan y) Which is (y-1) arctan y For the second integral we have 2*integral .5/(y-1) factor out .5 so we have 2*.5 * integral 1/(1-y) Which is integral 1/(1-y) which we see as ln(y-1) So bringing the 2 intervals together we get (y-1) arctan y + ln(y-1)+c confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Let's integrate just y / (y^3 - y^2 + y - 1), then double the result. The denominator factors by grouping: y^3 - y^2 + y 1 = (y^3 + y) (y^2 + 1) = y ( y^2 + 1) 1 ( y^2 + 1) = (y 1) ( y^2 + 1). Using partial fractions you would then have (a y + b) /(y^2 + 1) + c /(y-1) = y / ( (y^2+1)(y-1)) where we need to evaluate a, b and c. Putting the left-hand side over a common denominator (multiply the first fraction by (y-1)/(y-1) and the second by (y^2+1) / (y^2 + 1)) we obtain: [ (a y + b)(y-1) + c(y^2+1) ] / ((y^2+1)(y-1)) = y / ((y^2+1)(y-1)). The denominators are identical so the numerators are equal, giving us (a y + b)(y-1) + c(y^2+1) = y, or a y^2 + (-a + b) y - b + c y^2 + c = y. Grouping the left-hand side: (a + c) y^2 + (a - b) y + c - b = y. Since this must be so for all y, we have a + c = 0 (this since the coefficient of y^2 on the right is 0) -a + b = 1 (since the coefficient of y on the right is 1) c - b = 0 (since there is no constant term on the right). From the third equation we have b = c; from the first a = -c. So the second equation becomes c + c = 2, giving us 2 c = 2 so that c = 1. Thus b = c = 1 and a = -c = -1. Our integrand (a y + b) /(y^2 + 1) + c /(y-1) becomes (-y + 1 ) / (y^2 + 1) + 1/2 * 1 / (y-1), or -y / (y^2 + 1) + 1 / (y^2 + 1) + 1 / (y-1). An antiderivative is easily enough found with or without tables to be -1/2 ln(y^2 + 1) + arcTan(y) + ln | y - 1 | Doubling the result to get the integral of the given function we have -ln(y^2 + 1) + 2 arcTan(y) + 2 ln | y - 1 | + c, where c now stands for an arbitrary integration constant.
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Given Solution: If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2. So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to -u^.5. Translated in terms of the original variable z we get -sqrt(2z-z^2). If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2. So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to -u^.5. Translated in terms of the original variable z we get -sqrt(2z-z^2). DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: Section 7.4 Problem 9 7.4.9 (previously 7.4.36) partial fractions for 1 / (x (L-x))
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Integral 1/(x(L-x)) We use partial fractions So breaking apart we get a/x + b/(L-x) = 1/(x(L-x)) To get a common denominator we multiply the first term by (L-x) and the second term by x Getting for the numerators a(L-x) + bx =1 aL ax + bx = 1 x(-a+b) + aL=1 so aL=1 L=1/a and a=1/L -a+b=0 b= a so b=1/L so now we go back to our original expression integral a/x + b/(L-x) = 1/(x(L-x)) and substitute in Integral (1/L)/x + integral (1/L)/(L-x) 1/L integral (1/x) + 1/L integral 1/(L-x) 1/L lnx + 1/L ln(L-x) +c confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a / x + b / (L-x) = [ a (L-x) + bx ] / [ x(L-x)] = [ a L + (b-a)x ] / [ x(L-x)]. This is equal to 1 / [ x(L-x) ]. So a L = 1 and (b-a) = 0. Thus a = 1 / L, and since b-a=0, b = 1/L. The original function is therefore 1 / x + b / (L-x) = 1 / L [ 1 / x + 1 / (L-x) ]. Integrating we get 1 / L ( ln(x) - ln(L-x) ) = 1 / L ln(x / (L-x) ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### I am off somewhere and am confused on what I am doing incorrectly. Please let me know what I am not understanding?????
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Integral of (y+2)/(2y^2+3y+1) You can factor the denominator to get Integral (y+2)/[(2y+1)(y+1)] You can use #27 in the table but first you have to get the coefficients of the y values in the denominator to be the same by factoring out a 2 from the first term of the denominator getting 1/2integral (y+2)/[(y+.5)(y+1) Using # 27 letting a=-.5 b = -1 c=1 x=y d=2 ½[ 1 / (a - b) [ (ac + d) ln | x - a | - (bc + d) ln | x - b | ] + C.] ½[1/(.5+1) [(.5*1+2) ln|y+.5) |-(-1*1+2) ln|y+1| +c] ½*2(1.5ln|y+.5|-ln|y+1|+c] 1.5ln|y+.5|-ln|y+1|+c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (y+2) / (2y^2 + 3 y + 1) = (y + 2) / ( (2y + 1) ( y + 1) ) = (y + 2) / ( 2(y + 1/2) ( y + 1) ) = 1/2 * (y + 2) / ( (y + 1/2) ( y + 1) ) The expression (y + 2) / ( (y + 1/2) ( y + 1) ) is of the form (cx + d) / ( (x - a)(x - b) ) with c = 1, d = 2, a = -1/2 and b = -1. Its antiderivative is given as 1 / (a - b) [ (ac + d) ln | x - a | - (bc + d) ln | x - b | ] + C. The final result is obtained by substitution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: STUDENT COMMENTS: I have had some trouble figuring out this section. I couldn't figure out how to break down this denominator to make a partial fraction. Also I do not understand a step in the process in general. In the #10 class notes it explains another problem: This one is 1/[(x -3)(x +5)]. I understand this part of the notes: """"We know that when a fraction with denominator x-3 is added to a fraction with denominator x+5 we will obtain a fraction whose denominator is (x-3)(x+5). We conclude that it must be possible to express the given fraction as the sum A / (x-3) + B / (x+5), where A and B are numbers to be determined. Setting the original fraction equal to the sum we obtain an equation for A and B, as expressed in the third line. Multiplying both sides of the equation by (x-3) (x+5) we obtain the equation in the fourth line, which we rearrange to the form in the fifth line by collecting the x terms and the constant terms on the left-hand side and factoring."" This part I do not understand: """"Since the right-hand side does not have an x term, we see that A + B = 0"" How did you find that this equals 0? INSTRUCTOR RESPONSE: The equation for this function would be A / (x-3) + B / (x+5) = 1/[(x -3)(x +5)] To simplify the left-hand side need to obtain a common denominator. We multiply the first term by (x + 5) / (x + 5), and the second term by (x - 3) / (x - 3): A / (x-3) * (x + 5) / (x + 5) + B / (x+5) * (x - 3) / (x - 3) = 1/[(x -3)(x +5)] so that A ( x + 5) / ( (x - 3) ( x + 5) ) + B ( x - 3) / ( (x - 3) (x + 5) ) = 1/[(x -3)(x +5)] . Adding the fractions on the left-hand side: ( A ( x + 5) + B ( x - 3) ) / ( ( x - 3) ( x + 5) ) = 1/[(x -3)(x +5)] . Simplifying the numerator we have ( (A + B) x + (5 A - 3 B) ) / ( ( x - 3) ( x + 5) ) = 1/[(x -3)(x +5)]. The denominators are equal, so the equation is solved if the numerators are equal: (A + B) x + (5 A - 3 B) = 1. It is this last equation which lacks an x term on the right-hand side. To maintain equality the left-hand side must also have no x term, which can be so only if A + B = 0. The other term 5 A - 3 B is equal to 1. Thus we have the simultaneous equations A + B = 0 5 A - 3 B = 1. These equations are easily solve, yielding the solution A = 1/8, B = -1/8. CONTINUED STUDENT COMMENT: I understand this: """"we see that therefore 5A - 3B = 1, so we have two equations in two unknowns A and B."""" I could not figure out how you found A and B as shown below: Solving these equations we obtain B = -1/8, A = 1/8, as indicated. We conclude that the expression to be integrated is A / (x-3) + B / (x+5) = 1/8 * 1/(x-3) - 1/8 * 1/(x+5). INSTRUCTOR RESPONSE The system A + B = 0 5 A - 3 B = 1. can be solved by elimination or substitution. Using substitution: Solve the first equation for A, obtaining A = -B. Substitute this value of A into the second equation. obtaining 5 * (-B) + (-3 B) = 1 so that -8 B = 1 and B = -1/8. Go back to the fact that A = -B to obtain A = - (-1/8) = 1/8. To solve by elimination you could add 3 times the first equation to the second, eliminating B and obtaining 8 A = 1, so that A = 1/8. Substituting this back into the first equation we obtain 1/8 + B = 0 so that B = -1/8. "